The formation of the oxide ion, ${\mathrm{O}}_{\left(\mathrm{g}\right)}^{-2}$from oxygen atom requires first an exothermic andthen an endothermic step as shown below

 $$\begin{gathered} {\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{e}}^{-}\to {\mathrm{O}}_{\left(\mathrm{g}\right)}^{-};{\mathrm{\Delta e}}_{\mathrm{g}}{\mathrm{H}}_1=-141{\mathrm{kJmol}}^{-1} \\ {\mathrm{O}}_{\left(\mathrm{g}\right)}^{-}+{\mathrm{e}}^{-}\to {\mathrm{O}}_{\left(\mathrm{g}\right)}^{-2};{\mathrm{\Delta e}}_{\mathrm{g}}{\mathrm{H}}_1=+780{\mathrm{kJmol}}^{-1} \end{gathered}$$

Thus process of formation of O^–2 in gas phase is unfavourable even though O^–2 is isoelectronic with Neon. It is due to the fact that

Given, an exothermic step must come first, followed by an endothermic step, to produce the oxide ion, ${\mathrm{O}}^{2-}\left(\mathrm{g}\right)$, from an oxygen atom:

$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-}\left(\mathrm{g}\right);{\Delta }_{\mathrm{f}}{\mathrm{H}}^{\ominus }=-141 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{2-}\left(\mathrm{g}\right);{\Delta }_{\mathrm{f}}{\mathrm{H}}^{\ominus }=+780 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

Despite the fact that ${\mathrm{O}}^{2-}$and neon have identical electron configurations, the process of ${\mathrm{O}}^{2-}$formation in the gas phase is unfavorable. because the stability gained by achieving a noble gas configuration is outweighed by the electron repulsion.

The electrostatic attraction between the two negative charges is very strong when an electron is added to an ${\mathrm{O}}^{-}$ anion. As a result of this, it is observed that the second electron gain enthalpy is positive for oxygen. 

Therefore, option(C) is the answer.